Question

Evaluate this integral
$\bf{ 1 \atop \huge \int _{ \small \atop0}} x(tan^{ - 1} x^{2} )dx$
Please answer only If you know the answer.​

Answer 1

♣ Qᴜᴇꜱᴛɪᴏɴ :

Evaluate the given integral :

$\boxed{\sf{\int _0^1\:x\left(tan^{-1}x^2\right)dx}}$

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♣ ᴀɴꜱᴡᴇʀ :

$\boxed{\sf{\int _0^1\:x\left(tan^{-1}x^2\right)dx=\dfrac{\pi }{8}-\dfrac{1}{4}\ln \left(2\right)\quad }}$

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♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\large\boxed{\sf{\int _0^1\:x\arctan \left(x^2\right)dx}}$

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$\bf{Step\:\:1:\ Apply\:\: Integration\:\: By \:\:Parts}$

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Integration By Parts definition :

$\sf{\int \:uv'=uv-\int \:u'v}$

$\sf{u=\arctan \left(x^2\right)}$

$\sf{v'=x}$

$\underline{\underline{u^{\prime}=\dfrac{d}{d x}\left(\arctan \left(x^{2}\right)\right)}}$                                                

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Steps :

$\text { Apply the chain rule: } \dfrac{1}{\left(x^{2}\right)^{2}+1} \dfrac{d}{d x}\left(x^{2}\right)$

$=\dfrac{1}{\left(x^2\right)^2+1}\dfrac{d}{dx}\left(x^2\right)$

$=\dfrac{1}{\left(x^2\right)^2+1}\cdot \:2x$

$\sf{=\dfrac{2x}{x^4+1}}$

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$\underline{\underline{v=\int x d x}}$

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Steps :

$\text { Apply the Power Rule: } \int x^{a} d x=\dfrac{x^{a+1}}{a+1}, \quad a \neq-1$

$=\dfrac{x^{1+1}}{1+1}$

$=\dfrac{x^2}{2}$

$\text { Add a constant to the solution }$

$\sf{=\dfrac{x^2}{2}+C}$

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$\sf{\int \:uv'=uv-\int \:u'v}$

$\sf{=\left[\arctan \left(x^2\right)\dfrac{x^2}{2}-\int \dfrac{2x}{x^4+1}\cdot \dfrac{x^2}{2}dx\right]^1_0}$

$=\left[\dfrac{x^2\arctan \left(x^2\right)}{2}-\int \dfrac{2x}{x^4+1}\cdot \dfrac{x^2}{2}dx\right]^1_0$

$=\left[\dfrac{x^2\arctan \left(x^2\right)}{2}-\int \dfrac{x^3}{x^4+1}dx\right]^1_0$

$\sf{=\left[\dfrac{1}{2}x^2\arctan \left(x^2\right)-\int \dfrac{x^3}{x^4+1}dx\right]^1_0}$

Solve $\int \dfrac{x^3}{x^4+1}dx$

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Steps :

$\text { Apply u - substitution: } u=x^{4}+1$

$=\int \dfrac{1}{4u}du$

$\begin{aligned}&\text { Take the constant out: } \int a \cdot f(x) d x=a \cdot \int f(x) d x\\&=\dfrac{1}{4} \cdot \int \dfrac{1}{u} d u\end{aligned}$

$\begin{aligned}&\text { Use the common integral: } \int \dfrac{1}{u} d u=\ln (|u|)\\&=\dfrac{1}{4} \ln |u|\end{aligned}$

$\begin{array}{l}\text { Substitute back } u=x^{4}+1 \\\\=\dfrac{1}{4} \ln \left|x^{4}+1\right|\end{array}$

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$\sf{=\left[\dfrac{1}{2}x^2\arctan \left(x^2\right)-\dfrac{1}{4}\ln \left|x^4+1\right|\right]^1_0}$

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$\bf{Step\:\:2\:: Compute\:\: the\:\: boundaries}$

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$\sf{\int _a^bf\left(x\right)dx=F\left(b\right)-F\left(a\right)=\lim _{x\to \:b-}\left(F\left(x\right)\right)-\lim _{x\to \:a+}\left(F\left(x\right)\right)}$

$\lim _{x \rightarrow 0+}\left(\dfrac{1}{2} x^{2} \arctan \left(x^{2}\right)-\dfrac{1}{4} \ln \left|x^{4}+1\right|\right)$

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Steps :

$\text { Plug in the value } x=0$

$=\dfrac{1}{2}\cdot \:0^2\arctan \left(0^2\right)-\dfrac{1}{4}\ln \left|0^4+1\right|$

$=0$

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$\lim _{x\to \:1-}\left(\dfrac{1}{2}x^2\arctan \left(x^2\right)-\dfrac{1}{4}\ln \left|x^4+1\right|\right)$

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Steps :

$\text { Plug in the value } x=1$

$=\dfrac{1}{2}\cdot \:1^2\arctan \left(1^2\right)-\dfrac{1}{4}\ln \left|1^4+1\right|$

$=\dfrac{\pi }{8}-\dfrac{1}{4}\ln \left(2\right)$

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$\sf{\int _a^bf\left(x\right)dx=F\left(b\right)-F\left(a\right)=\lim _{x\to \:b-}\left(F\left(x\right)\right)-\lim _{x\to \:a+}\left(F\left(x\right)\right)}$

$=\dfrac{\pi }{8}-\dfrac{1}{4}\ln \left(2\right)-0$

$\large\boxed{\sf{=\dfrac{\pi }{8}-\dfrac{1}{4}\ln \left(2\right)}}$